Simple Polynomial

Use a trial and error method. Find a value of 'x' that would make f(x)=0 for some value of 'a'. Hint: what's f(1), and what value of 'a' corresponds?

 

If x₀ is a double root and x₁ is a different root, then your polynomial can be written 2(x- x₀)²(x- x₁). Multiply that out compare to the given polynomial.

"Multiple root" might mean a triple root: 2(x- x₀)³. Multiply that out and compare to the given polynomial.

 

Have you tried Rational Roots Theorem and synthetic division? Picking rational roots to try will not immediately involve 'a'. The synthetic division will certainly involve 'a'. The use of synthetic division either gives remainder of zero for a root or a nonzero remainder if a choice is not a root.

 

The trouble with rational roots and factoring etc, is that the double root doesn't look like it's rational. Try this. If x is a double root then f(x)=0 and f'(x)=0. Solve those simultaneously. You wind up having to solve a cubic.

 

What is repeated factor? Does it mean that it has two or more same roots? (Its translation problem)

 

Yes, it has two or more of the same root.

 

Why would you think that this equation will have a rational root?